Set 62 (Answers begin on page 274)
Some common errors in solving equations and
inequalities, as well as simplifying algebraic expressions,
are explored in this problem set.
961.The solution set for the inequality –6x24 is
(–4,∞).
a.The inequality sign must be switched when
multiplying both sides by a negative real
number. The correct solution set should be
(∞,–4).
b.You should multiply both sides by –6, not
divide by –6. The correct solution set should
be (–144,∞).
c.There is no error.
962.The solution set for the equation |x– 1| = 2
is {–1}.
a.There are two solutions of this equation,
namely x= –1 and x= 3.
b.The solution of an absolute value equation
cannot be negative. The only solution is x= 3.
c.There is no error.
- log 3 1 = 0
a.1 is an invalid input for a logarithm. As such,
the quantity log 3 is undefined.
b.The input and output are backward. The real
statement should read log 3 0 = 1.
c.There is no error.
964.The solution of –x+^4 7 = xx++^37 is x= –7.
a.The equation obtained after multiplying
both sides by x+ 7 was not solved correctly.
The correct solution should be x= 1.
b.x= –7 cannot be the solution because it
makes the terms in the original equation
undefined—you cannot divide by zero. As
such, this equation has no solution.
c.There is no error.965.The solutions of the equation log 5 x+ log 5 (5x^3 )
= 1 are x= –1 and x= 1.
a.Both solutions should be divided by 5 ;
that is, the solutions should be x= .
b.While x= 1 satisfies the original equation,x
= –1 cannot because negative inputs into a
logarithm are not allowed.
c.There is no error.966.The complex solutions of the equation x^2 + 5 = 0
obtained using the quadratic formula are given
byx== 2 i 5 .
a.The denominator in the quadratic formula is
2 a,which in this case is 2, not 1. As such, the
complex solutions should be x =i 5 .
b.There are no complex solutions to this equa-
tion because the graph ofy= x^2 + 5 does not
cross the x–axis.
c.There is no error.967.x^2 – 4x– 21 = (x+ 7)(x– 3)
a.This is incorrect because multiplying the
binomials on the right side of the equality
yields x^2 – 21, which is not the left side
listed above.
b.The signs used to define the binomials on
the right side should be switched. The cor-
rect factorization is (x– 7)(x + 3).
c.There is no error.^0 ^0 ^2 – 4(1)(5)
1^1
5 - COMMON ALGEBRA ERRORS–